Well, nice property of the non-squarefree numbers . Here's a sketch of the official sol. So note that if $n=ab+bc+ca$, then $n+a^2=(a+b)(a+c)$, so we have to construct $a$ for each non-squarefree $n$, such that $n+a^2$ is representable as the product of two numbers, bigger than $a$. Consider prime $p$, such that $n=p^2l$. So firstly consider whether we can take $a=p$. We want $(l+1)p^2$ to represented as a product in the above way. If $l+1>p$, we have It. If $l+1$ is composite, then it is $st$, and take $ps$ and $pt$. So we are left with the case when it's prime $q$, so $n=(q-1)p^2$. Now take $p=mq+r$, where $r$ is the remainder (now we look in the case where $r$ is positive integer) , and choose $a=r$ and the rest is to choose the one number to be $q>r$, the other is bashed to be bigger than $r$. We are only left to consider $q=p$, then choose $a=6$ and factor out $p^3-p^2+36$ to see that it works.

put $c=1$ . $$ \rightarrow ab+a+b = a(b+1)+b = n $$$$ a= \dfrac {n-b} {b+1} $$So we just need to find $b$ such that $b+1 | n-b $ . And also $b+1 | b+1$ , so $b+1 | n+1$ . Clearly we can find a positive integer $b$ such that the condition become true. After finding $b$ , $a$ is determined too from the fraction.

E.A.K wrote: put $c=1$ . $$ \rightarrow ab+a+b = a(b+1)+b = n $$$$ a= \dfrac {n-b} {b+1} $$So we just need to find $b$ such that $b+1 | n-b $ . And also $b+1 | b+1$ , so $b+1 | n+1$ . Clearly we can find a positive integer $b$ such that the condition become true. After finding $b$ , $a$ is determined too from the fraction. THere are some problems in your solution if $n+1$ is prime.

Pick any prime $p \mid k$ and write $n = xp^2$ with $x \in \mathbb Z_{>0}$. If $x \ge p$, then take $b=p,c=p(p-1)$. Then $$ a = \frac{n - p^2(p-1)}{p^2} \in \mathbb Z_{>0} $$works. Assume $x \le p-1$ now. As $n > 18$, so this forces $p \ge 5$. If $x = p-1$, take $a=6$. We require $$ (6+b)(6+c) = (a+b)(a+c) = ab + bc + ca + a^2 = n + 36 = p^2(p-1) + 36 = (p+3)(p^2 - 4p + 12) $$As $p \ge 5$, so both $p+3,p^2 - 4p + 12$ are $>6$. Hence $b,c$ can be positive integers. Lastly assume $x \le p-2$. Observe $x+1 \mid n + p^2$. Pick $0 < r < k$ with $r \equiv p \pmod{x+1}$ (recall $x+1 \nmid p$). Take $a=r$. We require $$ (r+b)(r+c) = (a+b)(a+c) = n + r^2 = (x+1) \cdot \frac{n + r^2}{x+1} $$Now $x+1 > r$ as $r < x$. Also, $\frac{n+r^2}{x+1} > r$ since $n \ge p^2 > (x+1)^2$. Hence $b,c$ can be positive integers. $\blacksquare$

We need to get $n+a^2=(a+b)(a+c)=xy$ $(\diamondsuit)$ for $x,y\in \mathbb{Z}^+$ and $x,y>a.$ Let $n=p^2m$ for prime $p$ and $m\in \mathbb{Z}^+.$ Case $m+1>p$. Plugging $a=p$ we get $n+a^2=p^2\cdot (m+1)$ which satisfies $(\diamondsuit)$. Case $m+1=p$. Since $n=p^3-p^2>18$ we may assume $p\geq 5.$ Plugging $a=6$ we get $n+a^2=(p+3)(p^2-4p+12)$ which satisfies $(\diamondsuit)$. Case $m+1<p$. If $m+1$ is a composite number $st$ for integers $s,t>1$ we put $a=p$ to get $n+a^2=ps\cdot pt$. If $m+1$ is a prime, assume that $p=l(m+1)+r$ for a remainder $r$ of $p$ modulo $m+1.$ Plugging $a=r$ we get $n+a^2=(m+1)(l^2(m+1)+2lr+r^2)$ which satisfies $(\diamondsuit)$. This finishes our proof.

If $n=ab+bc+ac$ then $n+a^2=(a+b)(a+c)$ so we should prove that for any $n$ non-squarefree exists $a$ such that : $n+a^2=xy , x,y>a$ Case $m+1>p$. Plugging $a=p$ we get $n+a^2=p^2\cdot (m+1)$ which satisfies. Case $m+1=p$. Since $n=p^3-p^2>18$ we may assume $p\geq 5.$ Plugging $a=6$ we get $n+a^2=(p+3)(p^2-4p+12)$ which satisfies. Case $m+1<p$. If $m+1$ is a composite number $st$ for integers $s,t>1$ we put $a=p$ to get $n+a^2=ps\cdot pt$. If $m+1$ is a prime, assume that $p=l(m+1)+r$ for a remainder $r$ of $p$ modulo $m+1.$ Plugging $a=r$ we get $n+a^2=(m+1)(l^2(m+1)+2lr+r^2)$ which satisfies . This finishes our proof.

Semyon's favorite factoring trick Rewrite the equation as $n+c^2=(c+a)(c+b)$, so we want to find $c$ such that $n+c^2$ can be written as the product of two numbers both greater than $c$. Let $n=p^2m$ where $p$ is minimal. If $p=2$ take $c=2$. If $m+1$ is composite we can take $c=p$ and write $n+c^2$ as the product of $xp$ and $yp$ where $x,y>1$ and $xy=m+1$. Otherwise if $m+1:=q$ is prime then we have $n=p^2(q-1)$. If $q>p$ let $c=p$ and note that both $p^2$ and $q$ are greater than $p$. Otherwise let $p=kq+r$ where $k,r\geq 0$ and $k$ is maximal. If $r>0$ let $c=r$, in which case $q \mid p^2(q-1)+r^2$. In this case we obviously have $q>r$, but also $$\frac{p^2(q-1)+r^2}{q}=\frac{(kq+r)^2(q-1)+r^2}{q}=k^2q^2-k^2q+2kqr-2kr+r^2=k(q-1)(k+2)+r^2>r.$$Otherwise $r=0$ so $p=q$, and $n=p^3-p^2$. Then let $c=6$; since $p^3-p^2+36=(p+3)(p^2-4p+12)$ and $p\geq 5$ otherwise $n$ is too small, both of these factors will be greater than $6$, so we are done. $\blacksquare$